I would just like so clarification is if possible in the words "Justify" in the question I have shown below:
So when they say justify I assume they mean show?, so my answers is no, the reason being is a follows:
Justification $$\hat{H}\bar{x}=\lambda\bar{x}\:[1]$$
Given $\bar{x}=\begin{pmatrix}1\\ 0\end{pmatrix}$
substituting this in to RHS of equation [2]
I get the following $$\begin{pmatrix}\epsilon _0&-\Delta \\ -\Delta &\epsilon _0\end{pmatrix}\begin{pmatrix}1\\ 0\end{pmatrix}=\begin{pmatrix}\epsilon _0\\ -\Delta \end{pmatrix}$$
So as the RHS of my equation [2] should be equal to my left of [2] which is not and I cannot factor anything from this equation as both are independent constants >0 it is shown that $|a\rangle$ is not an eigenstate.
Is this the type of justification they are requiring? I know you can you the determinant a calculate the eigenvalue and then the calculate the eigenstate that was but it seems a bit long winded.
In addition to the mathematical calculation, you can also offer a physical justification. Since it would not cost any energy to get from $|a>$ to $|b>$, it would be expected that the eigenstates are a linear combination of the two. If the Hamiltonian matrix is diagonal, your given states are eigenvectors. Since you have non-diagonal terms, those can be associated with a transition between the states. The two different eigenvalues, $\epsilon_0-\Delta$ and $\epsilon_0+\Delta$, correspond to eigenvectors $|a>+|b>$, and $|a>-|b>$.